FM numerical: deviation, index, 98% bandwidth, power

Given: $A_m = 10$ V, $f_m = 4000$ Hz, $A_c = 80$ V, $f_c = 10^7$ Hz = 10 MHz, $k_f = 400$ Hz/V.

a) Peak frequency deviation

$$\Delta f = k_fA_m = 400 \times 10 = 4000\ \text{Hz} = 4\ \text{kHz}$$

b) Modulation index

$$\beta = \frac{\Delta f}{f_m} = \frac{4000}{4000} = 1$$

c) Bandwidth containing over 98% of power

Carson's rule is based on keeping about 98% of the power:

$$B_T = 2(\Delta f + f_m) = 2(4 + 4) = 16\ \text{kHz}$$

Check with Bessel functions for $\beta = 1$: $J_0 = 0.7652$, $J_1 = 0.4401$, $J_2 = 0.1149$, $J_3 = 0.0196$.

So sidebands up to $n = 2$ are needed for more than 98% power:

$$B = 2nf_m = 2 \times 2 \times 4 = 16\ \text{kHz}$$

which agrees with Carson's rule.

d) Total power in unit impedance

FM has constant amplitude, so

$$P = \frac{A_c^2}{2R} = \frac{80^2}{2 \times 1} = 3200\ \text{W}$$

Answer: $\Delta f = 4$ kHz; $\beta = 1$; $B = 16$ kHz; $P = 3200$ W.